Material and Energy Balances: cram pack

Formula sheets, the key method steps for every topic and 8 practice problems, with full solutions at the back.

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1Formulas and methods

Mass balance

Steady state
in = out
Total balance
ṁ1 + ṁ2 = ṁ3
Component balance
ṁ1x1 + ṁ2x2 = ṁ3x3
Evaporator
F·xF = P·xP, W = F − P

Energy balance

Sensible heat
Q = ṁ·cp·ΔT
Latent heat
Q = ṁ·hfg
Mixing
ΣṁcpTin = ΣṁcpTout
Water
cp ≈ 4.18 kJ/(kg·K)

Ideal gas law

Ideal gas law
pV = nRT
Gas constant
R = 8.314 J/(mol·K)
Combined gas law
p1V1/T1 = p2V2/T2
Moles
n = m/M
Temperature
T[K] = T[°C] + 273.15

Stoichiometry

Moles
n = m/M
Concentration
c = n/V
Mole ratio
use the coefficients of the balanced equation
Limiting reactant
the one that runs out first
Percent yield
actual / theoretical × 100%

Method

  1. Write and balance the chemical equation.
  2. Convert the given mass to moles: n = m/M.
  3. Use the coefficients as the mole ratio to find the moles of the substance you want.
  4. Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.

2Practice problems

1.Mass balanceMedium
MixerStream 1600 kg/h · 45% NaOHStream 2ṁ₂ = ? · 10% NaOHProduct30% NaOH

How much of stream 2 must be added each hour to stream 1 so that the blended product contains 30% NaOH? Find ṁ2.

ṁ1 = 600 kg/hx1 = 45 %x2 = 10 %x3 = 30 %
ṁ2 = kg/h
2.Mass balanceHard
EvaporatorFeed F2400 kg/h · 7% NaOHProduct PP = ? · 65%Vapour W (pure water)0% NaOH

An evaporator concentrates a NaOH solution by boiling off pure water. Find the product flow P in kg/h.

F = 2400 kg/hxF = 7 %xP = 65 %
P = kg/h
3.Energy balanceMedium
hot sidecold sidelight oil · 170 °C70 °Cwater · 30 °Ct = ? °C

Hot light oil cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.

ṁh = 0.5 kg/scp,h = 2.1 kJ/(kg·K)Th: 170 → 70 °Cṁc = 3 kg/scp,c = 4.18 kJ/(kg·K)tin = 30 °C
tout = °C
4.Energy balanceHard
hot sidecold sideglycol mix · 120 °C60 °Cwater · 20 °C40 °C · ṁ = ?15% lost

Hot glycol mix heats water in an exchanger that loses 15% of the heat released by the hot stream to the surroundings. Find the water flow rate ṁc needed.

ṁh = 3 kg/scp,h = 3.6 kJ/(kg·K)Th: 120 → 60 °Ct: 20 → 40 °Ccp,c = 4.18 kJ/(kg·K)loss = 15 %
ṁc = kg/s
5.Ideal gas lawMedium
460 g of HeP = 200 kPaT = 50 °CV = ?

What volume V does 460 g of He occupy at 200 kPa and 50 °C? Treat it as an ideal gas.

m = 460 gM = 4.003 g/molP = 200 kPaT = 50 °CR = 8.314 J/(mol·K)
V = L
6.Ideal gas lawHard
State 1State 2

A fixed amount of CH₄ goes from 400 kPa, 2 L, 15 °C to 650 kPa and 180 °C. Find the new volume V2.

P1 = 400 kPaV1 = 2 LT1 = 15 °CP2 = 650 kPaT2 = 180 °C
V2 = L
7.StoichiometryMedium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

6 g of C3H8 and 16 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?

M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
8.StoichiometryHard
m(C6​H12​O6​)n(C6​H12​O6​)n(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 16 g of C6H12O6 reacts completely (the other reactants are in excess) and 7.2 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 7.2 g
Y = %

3Solutions

1.Mass balance

Answer: ṁ2 = 450 kg/h

Steady-state mass balance: in = out for the total and for each component. Blending problems combine both.

Balances

Total: ṁ3 = ṁ1 + ṁ2
NaOH: ṁ1x1 + ṁ2x2 = (ṁ1 + ṁ2)x3

Solve for ṁ2

ṁ2 = ṁ1(x1 − x3)/(x3 − x2) = 600·(45 − 30)/(30 − 10) = 450 kg/h

Answer

ṁ2 = 450 kg/h
2.Mass balance

Answer: P = 258.5 kg/h

Tie component: a component that passes through unchanged in one stream (here the solute) links the flows directly. Then the total balance gives the rest.

Solute balance

F·xF = P·xP → P = 2400·7/65 = 258.5 kg/h

Answer

P = 258.5 kg/h
3.Energy balance

Answer: tout = 38.37 °C

Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).

Heat released by the hot stream

Q = 0.5·2.1·(170 − 70) = 105 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 30 + 105/(3·4.18) = 38.37 °C

Answer

tout = 38.37 °C
4.Energy balance

Answer: ṁc = 6.589 kg/s

Energy balance with losses: heat released = heat absorbed + heat lost.

Heat released

Qh = 3·3.6·(120 − 60) = 648 kW

Heat that reaches the water

Qc = (1 − 0.15)·648 = 550.8 kW

Water flow rate

ṁc = Qc/(cp,c·Δt) = 550.8/(4.18·20) = 6.589 kg/s

Answer

ṁc = 6.589 kg/s
5.Ideal gas law

Answer: V = 1544 L

Ideal gas law: PV = nRT, with n = m/M.

Moles

n = 460/4.003 = 114.9 mol
T = 50 + 273.15 = 323.1 K

Volume

V = nRT/P = 114.9·8.314·323.1/200 = 1544 L

Answer

V = 1544 L
6.Ideal gas law

Answer: V2 = 1.936 L

Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.

Absolute temperatures

T1 = 288.1 K, T2 = 453.1 K

Combined gas law

V2 = V1·(P1/P2)·(T2/T1) = 2·(400/650)·(453.1/288.1) = 1.936 L

Answer

V2 = 1.936 L
7.Stoichiometry

Answer: m = 13.2 g

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Moles of each reactant

n(C3H8) = 6/44.1 = 0.1361 mol
n(O2) = 16/32 = 0.5 mol

Limiting reactant

0.1361/1 = 0.1361, 0.5/5 = 0.1 → O2

Mass of product

n(CO2) = 0.1·3 = 0.3 mol
m = 0.3·44.01 = 13.2 g

Answer

m = 13.2 g
8.Stoichiometry

Answer: Y = 87.99 %

Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.

Theoretical yield

n(C6H12O6) = 16/180.16 = 0.08881 mol
m = 0.08881·2/1·46.07 = 8.183 g

Percent yield

Y = 7.2/8.183·100 = 87.99%

Answer

Y = 87.99%

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